3 Single Particle Dynamics

3 Single particle dynamics 3.1 A rocket of initial mass mi is launched from the earth and accelerates until its fuel is exhausted. v The residual mass of the rocket's structure is mf . Assume that the fuel is burned at a constant posi- tive rate α and the exhaust gas is released vertically downward with a relative nozzle velocity u = const. Neglect air resistance, the decay of the gravitational acceleration g with altitude and all complications arising from the rotation of the earth. m(t) 1. Find an expression for the velocity of the g rocket v at time t. (a ) v(t) = − 1 gt2 + u2 ln mi 2 mi+αt (b ) v(t) = −gt + u ln mi mi−αt (c ) v(t) = −gt + u ln mi−αt mi (d ) v(t) = −gt + u (ln (mi) + ln (mi − αt)) 1 2 1 (e ) v(t) = − 2 gt + u ln 1−αt 2. Determine the altitude the rocket reaches when all the fuel is burned.u 2 1 mi−mf h mi−mf mf mf i (a ) z(tf ) = − g + u + ln 2 α α α mi 2 1 mi−mf mf mf (b ) z(tf ) = − g + u ln 2 α α mi mi−mf h mi−mf mf mf i (c ) z(tf ) = −g + u − ln α α α mi 2 h i 1 mi−mf mi−mf mf 2 mi (d ) z(tf ) = − g + u + ln 2 α α α mf 2 1 mi−mf h mi−mf mf mf i (e ) z(tf ) = − g + + ln 2 α α α mi Brief explanation of the answers - Question 1 We are asked to calculate the velocity of the rocket, therefore we can immediately exclude answer (a), (d) and (e) by looking at the units. By looking at the rocket equation, it can be seen that if the derivative of the mass with respect to time is negative, a force accelerating the rocket is generated (u is also negative.) Looking at the second term of answer (c), mi − αt will converge to mf , where mf < mi, i.e. the second term will be negative which implies that the velocity of the rocket is always decreasing while its losing mass. This is incorrect, leaving answer (b) as the correct answer. 3-1 3 Single particle dynamics 3-2 Solution It suffices to consider motion along the z-axis only. We denote the position of the rocket by r = zk, and at the initial time ti we pose z(ti) = 0. The velocity is given byr _ = v = vk, where v =z _ and v(ti) = 0. The exhaust velocity is u = −uk. For the external force we take F = mg = −mgk. The rocket equation reads: dv dm m + u = −mg : (3.1) dt dt By taking the initial time ti = 0, the expression for the mass of the rocket at time t is m − m m(t) = m − αt; if t ≤ t = i f : (3.2) i f α By dividing (3.1) by m and integrating both sides from 0 to t (with t ≤ tf ) and using the initial condition v(0) = 0 we get Z t u dm v(t) = −g − dτ: (3.3) 0 m dτ Using (3.2) andm _ = −α: Z t 1 v(t) = −gt + u α dτ: (3.4) 0 mi − ατ Substituting µ = mi − ατ, and accordingly dµ = −αdτ, Z m(t) 1 v(t) = −gt − u dµ. (3.5) mi µ The final result thus becomes m m v(t) = −gt + u ln i = −gt + u ln i : (3.6) m(t) mi − αt Brief explanation of the answers - Question 2 By inspection, answers (c), (d) and (e) cannot be correct as a result of having terms with inconsistent units. This leaves us with answer (a) and (b), where answer (b) is the result of an error in the derivation. It is concluded that answer (a) is correct. Solution The velocity of the rocket is given by m m v(t) = −gt + u ln i = −gt + u ln i (3.7) m(t) mi − αt 3 Single particle dynamics 3-3 mi−mf Integrating (3.6) from 0 to tf = α and using z(0) = 0: Z tf 1 2 αs z(tf ) = − gtf − u ln 1 − ds (3.8) 2 0 mi 1 2 mi αtf = − gtf + u tf − tf − ln 1 − (3.9) 2 α mi 1 m − m 2 m − m m m = − g i f + u i f + f ln f : (3.10) 2 α α α mi 3 Single particle dynamics 3-4 3.2 The strings AC and BC connect a ball of mass m to the vertical shaft, as shown. When the shaft rotates at the constant angular speed θ_, the ball travels in a A horizontal circle with the strings inclined at an angle α to the shaft. What is the magnitude of the tension force T AC ? ↵ " # 1 mRθ_2 mg T (a ) T = + AC AC 2 sin α cos α ✓˙ " # R 1 mRθ_2 mg m (b ) T = − AC 4 sin α cos α C g T BC 1 mRθ_2 ↵ (c ) T = AC 2 sin α 1 mg (d ) TAC = B 2 cos α (e ) TAC = 0 Solution Since the motion is curvilinear, it is convenient to refer to the cylindrical coordinate system depicted in the figure. The particle moves on a circle with constant T AC ez en et T BC mg velocity v = θR_ . Therefore, only a normal acceleration is present. The linear momentum principle, written in cylindrical components 1, gives _ _2 P = mRθ en = (TAC + TBC ) sin αen + ((TAC − TBC ) cos α − mg) ez: (3.11) By equating the components, we obtain 1Note that the same formal procedure of projection the linear momentum principle written for an inertial frame onto the moving coordinates must be carried out, as done in previous exercises. 3 Single particle dynamics 3-5 _2 mRθ = (TAC + TBC ) sin α (3.12) 0 = (TAC − TBC ) cos α − mg (3.13) which can be solved for TAC : " # 1 mRθ_2 mg T = + (3.14) AC 2 sin α cos α 3.3 A block of mass m slides down a frictionless in- y1 cline, which translates with constant acceleration aex. m The angle of inclination is α. At t = 0, the incline is h a gg x ↵ at rest. What is the correct equation of motion of the 2 ↵ system? x1 2 (a )y ¨1 = g sin α − a cos α sin α (b )y ¨1 = g sin α − a cos α (c )y ¨1 = g − a cos α sin α (d )y ¨1 = g sin α (e )y ¨1 = 0 Solution The coordinate transformation between the inertial frame and the moving frame is 1 x = at2 + y cos α (3.15) 1 2 1 x2 = h − y1 sin α: (3.16) The linear momentum principle in the inertial frame writes mx¨1 = N sin α (3.17) mx¨2 = −mg + N cos α; (3.18) where N is the magnitude of the normal force exerted on the block by the incline. Sub- stituting (3.15), we obtain ma + my¨1 cos α = N sin α (3.19) −my¨1 sin α = −mg + N cos α; (3.20) and, eliminating N and rearranging, we obtain y¨1 = g sin α − a cos α; (3.21) 3 Single particle dynamics 3-6 where m has been dropped, being a common factor. Note that the projection of the linear momentum principle on the accelerating frame generates an extra term −a cos α, which is not a physical force, but rather is a by-product of the fact that we are referring to a non-inertial frame of reference (in this case, linearly accelerating w.r.t. an inertial frame). Note also that, it is not possible to write the linear momentum principle in the non-inertial frame directly, as it simply does not hold for non-inertial frames. 3.4 A particle of mass m is at rest in the horizontal (x ; x ) inertial frame. An observer is fixed to 1 2 y2 the rotating (y1; y2) frame whose angular velocity is x2 θ_ = cos(t). No active force is applied to the particle. Which of the following statements is correct? m (a ) The observer in the (y ; y ) frame observes a 1 2 y y1 nonzero Euler force on the particle. ✓(t) (b ) The observer in the (y1; y2) frame observes a zero Coriolis force on the particle. x1 (c ) The observer in the (y1; y2) frame observes a nonzero active force on the particle. (d ) The reaction forces on the particle counteract the inertial forces arising from the rotation of the non-inertial frame. (e ) The centrifugal force and the Euler force are parallel to each other. Solution The reference frame (y1; y2) is a rotating frame, therefore it is a non-inertial frame. Since the motion is planar, we can directly use the linear momentum principle equation for non-inertial, rotating frames in 2D: my¨ = QT F + mθ_2y − 2mθk_ × y_ − mθk¨ × y (3.22) where k is the unit vector perpendicular to the (y1; y2) plane, and the terms of the right- hand side are respectively: 1. the active forces expressed in the moving coordinate frame (via the transformation matrix Q), 2. the centrifugal force, 3. the Coriolis force, 4. the Euler force. These last three terms are not active forces, but they arise from the differentiation in time of the transformation between the inertial and non-inertial frame of reference. We refer to these forces as inertial forces.

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